Solve P^2 [.1 .9] .6 .4
square of a matrix?
yes
Matrices..I totally forgot how to do this..
\[\mathbf A=\begin{bmatrix}.1,&.9\\.6,&.4\end{bmatrix}\] \[\mathbf A^2=\color{teal}{\mathbf A}\color{orange}{\mathbf A}=\color{teal}{\begin{bmatrix}.1,&.9\\.6,&.4\end{bmatrix}}\color{orange}{\begin{bmatrix}.1,&.9\\.6,&.4\end{bmatrix}}\\ =\begin{bmatrix}(\color{teal}{.1}\times \color{orange}{.1})+(\color{teal}{.9}\times\color{orange}{.6}),&(\color{teal}{.1}\times\color{orange}{.9})+(\color{teal}{.9}\times\color{orange}{.4}) \\(\color{teal}{.6}\times\color{orange}{.1})+(\color{teal}{.4}\times\color{orange}{.6}),&(\color{teal}{.6}\times\color{orange}{.9})+(\color{teal}{.4}\times\color{orange}{.4})\end{bmatrix}\]
Got it.. xD
pardon?
Do you understand @guruguru
@UnkleRhaukus I mean i had forgotten how to do this..but now i know.
ohh i got it @uri and Thanks @UnkleRhaukus for ur help :)
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